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%I #14 Nov 23 2021 01:37:43
%S 1,1,8,135,3544,126980,5778606,319234454,20755549256,1552791269232,
%T 131408062049040,12411898074678432,1294418444771718168,
%U 147733436055601473168,18315901821846419101416,2451257290708213030681080,352217918432527724627871936,54082428426583359310449351168
%N E.g.f. A(x) satisfies: A(x)^A(x) = 1/(1 - x*A(x)^4)
%C Generally, for A(x)^A(x) = 1/(1-x*A(x)^p) is limit n->infinity a(n)^(1/n)/n = exp(p*(1-r)/(r-p))*(p-r+exp(r/(p-r))), where r is the root of the equation exp(r/(p-r)) = (r-p)/r*(r + LambertW(-1,-r*exp(-r))
%C Generally, if e.g.f. A(x) satisfies A(x)^A(x) = 1/(1-x*A(x)^p), then a(n) ~ s*sqrt((s^s-1)/(p*(s^s-1)*(p*s^s-1)-s)) * n^(n-1) * (s^(p+s)/(s^s-1))^n / exp(n), where s is the root of the equation (1+log(s))*s = (s^s-1)*p. Compared with my previous result, limit n->infinity a(n)^(1/n)/n = s^(p+s)/(s^s-1)/exp(1). - _Vaclav Kotesovec_, Dec 28 2013
%H Seiichi Manyama, <a href="/A229237/b229237.txt">Table of n, a(n) for n = 0..326</a>
%F Limit n->infinity a(n)^(1/n)/n = exp(4*(1-r)/(r-4))*(4-r+exp(r/(4-r))) = 3.635561077783029..., where r = 0.8373821681637... is the root of the equation exp(r/(4-r)) = (r-4)/r*(r + LambertW(-1,-r*exp(-r))
%F a(n) ~ s*sqrt((s^s-1)/(4*(s^s-1)*(4*s^s-1)-s)) * n^(n-1) * (s^(4+s)/(s^s-1))^n / exp(n), where s = 1.3031377498774256189193761312... is the root of the equation (1+log(s))*s = 4*(s^s-1). - _Vaclav Kotesovec_, Dec 28 2013
%t Table[Sum[(4*n-k+1)^(k-1)*(-1)^(n-k)*StirlingS1[n,k],{k,0,n}],{n,0,20}]
%t p=4; E^(p*(1-r)/(r-p))*(p-r+E^(r/(p-r)))/.FindRoot[E^(r/(p-r))==(r-p)/r*(r+LambertW[-1,-r*E^(-r)]), {r,1/2}, WorkingPrecision->50] (* program for numerical value of the limit n->infinity a(n)^(1/n)/n *)
%Y Cf. A141209, A216135, A216136, A349561.
%K nonn
%O 0,3
%A _Vaclav Kotesovec_, Sep 17 2013
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